The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, λ12, of the photons emitted in this process is:

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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)
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  1. 20/7
  2. 27/5
  3. 7/5
  4. 9/7

Answer (Detailed Solution Below)

Option 1 : 20/7
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JEE Main 04 April 2024 Shift 1
90 Qs. 300 Marks 180 Mins

Detailed Solution

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Concept:

From question, the atom jumps from one spectral line to another spectral line. The wavelength of the atom is given by the formula:

Where,

‘Z’ is the atomic number

n1 and n2 is the principal quantum number

R is the Rydberg constant. (1.09737 × 107 m-1)

Calculation:

When the atom jumps from third excited state to the second excited state:

When the atom jumps from second excited state to the first excited state:

 

 

From question,

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